To determine the momentum after the collision, we need to consider the momentum vectors of both objects.
- The first object is traveling northeast with a momentum of \(12 , \text{kg-m/s}\).
- The second object is traveling southwest with a momentum of \(4 , \text{kg-m/s}\).
To find the total momentum after the collision, we can break this down into components:
-
Northeast momentum (first object):
- This can be split into x and y components (assuming northeast corresponds to a 45-degree angle from both axes):
- \( p_{1x} = 12 \cos(45^\circ) = 12 \cdot \frac{1}{\sqrt{2}} = 6\sqrt{2} , \text{kg-m/s} \)
- \( p_{1y} = 12 \sin(45^\circ) = 12 \cdot \frac{1}{\sqrt{2}} = 6\sqrt{2} , \text{kg-m/s} \)
-
Southwest momentum (second object):
- This also corresponds to a 45-degree angle but in the opposite direction:
- \( p_{2x} = -4 \cos(45^\circ) = -4 \cdot \frac{1}{\sqrt{2}} = -2\sqrt{2} , \text{kg-m/s} \) (negative because it's moving in the negative x direction)
- \( p_{2y} = -4 \sin(45^\circ) = -4 \cdot \frac{1}{\sqrt{2}} = -2\sqrt{2} , \text{kg-m/s} \) (negative because it's moving in the negative y direction)
Now, we can find the total momentum components in the x and y directions:
-
Total x-component of momentum: \[ p_{x, \text{total}} = 6\sqrt{2} + (-2\sqrt{2}) = (6 - 2)\sqrt{2} = 4\sqrt{2} , \text{kg-m/s} \]
-
Total y-component of momentum: \[ p_{y, \text{total}} = 6\sqrt{2} + (-2\sqrt{2}) = (6 - 2)\sqrt{2} = 4\sqrt{2} , \text{kg-m/s} \]
Now we can find the magnitude of the total momentum vector: \[ p_{\text{total}} = \sqrt{(p_{x, \text{total}})^2 + (p_{y, \text{total}})^2} = \sqrt{(4\sqrt{2})^2 + (4\sqrt{2})^2} = \sqrt{32 + 32} = \sqrt{64} = 8 , \text{kg-m/s} \]
To find the direction, since both components are positive, the direction of the total momentum is northeast.
Thus, the momentum after their collision is:
8 kg-m/s northeast.